JEE Main 2026 Jan 23 Shift 1, Mathematics Q16: Trigonometric Equations

Use $\cos 2\theta =2{\cos }^2\theta -1$:

$\sqrt{3}(2{\cos }^2\theta -1)+8\cos \theta +3\sqrt{3}=0$

$2\sqrt{3}{\cos }^2\theta +8\cos \theta +2\sqrt{3}=0$

Divide by 2: $\sqrt{3}{\cos }^2\theta +4\cos \theta +\sqrt{3}=0$, factoring as $(\sqrt{3}\cos \theta +1)(\cos \theta +\sqrt{3})=0$.

Reject $\cos \theta =-\sqrt{3}$ (impossible). Keep $\cos \theta =-\frac{1}{\sqrt{3}}$.

The interval $[-3\pi ,2\pi ]$ has length $5\pi$, i.e. 2.5 periods. The equation $\cos \theta =c$ (with $|c|<1$) has 2 solutions per period, plus an endpoint check. Direct enumeration gives 5 solutions.